Answer to Question #12448 in Calculus for Sharon

Question

Use L'Hopital's rule to find limit x(pi/2 -tan^-1x) as x approaches infinity

Answer 1

limit x(pi/2 -tan^-1x)_{x approaches infinity}=limit (pi/2 -tan^-1x)/(1/x)_{x approaches infinity}
Both functions 1/x and (pi/2 -tan^-1x) tend to 0 when x tends to infinity/ So we use L'Hopital's rule. (1/x)'=-1/x^2
(pi/2 -tan^-1x)'=-1/(1+x^2)
limit (pi/2 -tan^-1x)/(1/x)_{x approaches infinity}=limit {-1/(1+x^2)/(-1/x^2)}_{x approaches infinity}=
=limit {x^2/(1+x^2)}_{x approaches infinity}=1
To know more about L'Hopital's rule you can watch our video which contains examples and detailed explanations. Take a look!